CHEMICAL KINATICS

Question

when sulphur (in terms of S8)is heated at temp t ,at equlibrium the pressure of S8 falls by 30% from 1 atm, because s8 is partillay converted into s2 finf kp for this reactoin options A 2.96 B 6.14 C 204.8 D 0.54

Answer 1

S₈ ===> 4S₂

1 atm ....0 atm  initial

0.7 atm .. 1.2 atm  change

0.7 atm ...1.2 atm  equilibrium

K_p = P(S₂)⁴/P(S₈) 

Kp = (1.2)⁴/0.7

Kp = 2.96 Answer A

Answer 2

Balanced equation for the reaction

 

S₈ ↔ 4S₂

1       0       Initial

-x      4x     Change

1-x    4x     Equilibrium

Given

x = 0.3 atm

 

k_p = (P_S2)⁴/P_S8

    = (4x)⁴/(1 -x) = (1.2)⁴/(0.7)

    = 2.96